Question

How do use the first derivative test to determine the local extrema `f(x) = (x+1)(x-3)^2`?

Answer 1

Have a look:

Explanation:

You can use the Product and Chain Rule to find the derivative:
`f'(x)=1*(x-3)^2+(x+1)*2(x-3)=x^2-6x+9+2(x^2-3x+x-3)=`
`=x^2-6x+9+2(x^2-2x-3)=x^2-6x+9+2x^2-4x-6=`
`=3x^2-10x+3`
Now you can set it equal to zero to find the x coordinate(s) of the local extrema:
`3x^2-10x+3=0`
We can use the Quadratic Formula:
`x_(1,2)=(10+-sqrt(100-36))/6=(10+-8)/6=`
So you get:
`x_1=3` giving `y=0`
`x_2=1/3` giving `y=9.5`

Graphically:
graph{(x+1)((x-3)^2) [-10, 10, -5, 5]}